Problem packetResearch packetR819
Unary one-way NFAs have a tight quadratic two-way determinization cost
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The record cites sources for its explanation.
Recorded status: reported
Recorded scope: conversion of unary n-state one-way nondeterministic finite automata into equivalent two-way deterministic finite automata
Complete recorded scope and conditions
{
"kind": "family",
"statement": "conversion of unary n-state one-way nondeterministic finite automata into equivalent two-way deterministic finite automata",
"family": "unary 1NFAs"
}Originating problem: Polynomial determinization of two-way finite automata
Recorded relationships: The fixed-alphabet determinization question remains open
Authored record and scope
- Authored title
- Unary one-way NFAs have a tight quadratic two-way determinization cost
- Record type
- claim
- Stored status
- reported
- Evidence grade
- sourced
- Recorded scope data
- { "kind": "family", "statement": "conversion of unary n-state one-way nondeterministic finite automata into equivalent two-way deterministic finite automata", "family": "unary 1NFAs" }
- Linked research record IDs
- R815
2Authored explanation
Chrobak proves both directions in Section 6. Theorem 6.2 converts any unary \(n\)-state 1NFA to a 2DFA with \(O(n^2)\) states by first placing the NFA in unary normal form, checking the bounded prefix, and using separate passes for the eventual cycles. Theorem 6.3 gives an \(n\)-state unary 1NFA family for which every equivalent 2DFA needs \(\Omega(n^2)\) states.
This settles the tradeoff for one-way nondeterministic source automata over the singleton alphabet. Unary 2NFA determinization and the full fixed-alphabet 2NFA problem remain open. The 2003 erratum is recorded alongside the original article.
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3Evidence
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Verification source: doi.org ↗, Chrobak, Section 6, Theorems 6.2 and 6.3, pp. 156-157
4What was measured
5How it connects
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- claim
Recorded for
- problem
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"title": "Unary one-way NFAs have a tight quadratic two-way determinization cost",
"summary": "Every unary n-state 1NFA has an equivalent 2DFA with O(n^2) states, and some unary n-state 1NFAs require Omega(n^2) states in every equivalent 2DFA.",
"relevance": "For Polynomial determinization of two-way finite automata, record twnfa-claim-unary-one-way-tight-quadratic (“Unary one-way NFAs have a tight quadratic two-way determinization cost”) records a bound, answer, status fact, or structural consequence. The record states: Every unary n-state 1NFA has an equivalent 2DFA with O(n^2) states, and some unary n-state 1NFAs require Omega(n^2) states in every equivalent 2DFA.",
"relevance_source": "recorded",
"body": "Chrobak proves both directions in Section 6. Theorem 6.2 converts any unary \\(n\\)-state 1NFA to a 2DFA with \\(O(n^2)\\) states by first placing the NFA in unary normal form, checking the bounded prefix, and using separate passes for the eventual cycles. Theorem 6.3 gives an \\(n\\)-state unary 1NFA family for which every equivalent 2DFA needs \\(\\Omega(n^2)\\) states.\n\nThis settles the tradeoff for one-way nondeterministic source automata over the singleton alphabet. Unary 2NFA determinization and the full fixed-alphabet 2NFA problem remain open. The 2003 erratum is recorded alongside the original article.",
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"statement": "conversion of unary n-state one-way nondeterministic finite automata into equivalent two-way deterministic finite automata",
"family": "unary 1NFAs"
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"citation": {
"url": "https://doi.org/10.1016/0304-3975(86)90142-8",
"locator": "Chrobak, Section 6, Theorems 6.2 and 6.3, pp. 156-157"
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"source": {
"url": "https://doi.org/10.1016/0304-3975(86)90142-8",
"locator": "Chrobak, Section 6, Theorems 6.2 and 6.3, pp. 156-157"
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{
"slug": "R815",
"title": "The fixed-alphabet determinization question remains open",
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{
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}7Provenance
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